Assume I have a Jordan curve $C \subset \mathbb{R}^2$. Then by Schoenflies Theorem there exists a homeomorphism $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ such that $f(C)$ is the unit circle. Is there a simple way to go about finding one given a parametrized curve in the plane?

  • $\begingroup$ I presume you mean a homeomorphism of $\mathbb R^2$? $\endgroup$ – Jonas Meyer Sep 4 '11 at 21:37
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    $\begingroup$ This question seems somewhat related: math.stackexchange.com/questions/43796 $\endgroup$ – t.b. Sep 4 '11 at 21:38
  • $\begingroup$ Yes. Sorry I wasn't more specific. $\endgroup$ – Tim Seguine Sep 4 '11 at 21:38
  • $\begingroup$ @Theo Thanks for the link. I don't know why I didn't think to search for the Riemann mapping theorem. $\endgroup$ – Tim Seguine Sep 4 '11 at 21:40
  • $\begingroup$ Wouldn't such a simple method, if it existed, almost constitute a simple proof of the Jordan Curve Theorem? $\endgroup$ – hmakholm left over Monica Sep 4 '11 at 21:41

There are constructive versions of the Schoenflies theorem, but of course you have to be more precise about your input data.

For example, if you give a piecewise-linear simple closed curve in the plane, then there's a constructive version of the Riemann Mapping Theorem that applies. I believe the proof is fairly old. I'd look in Stephenson's Circle Packings book: http://www.math.utk.edu/~kens/

If your curve is smooth, then there's a less finitary but still fairly explicit way to construct the diffeomorphism. The idea would be to use the Gage-Hamilton curve straightening differential equation. This will turn your smooth curve into the round circle. The isotopy extension theorem would then provide your diffeomorphism.

Of course, there's Alexander's proof of the theorem. It's yet less explicit but it works quite well. The idea is that your curve in the plane divides the plane into an "inside" and "outside". A general linear function $f : \mathbb R^2 \to \mathbb R$ is a Morse function on the "inside" region. If all the critical points are at different "heights" this Morse function divides the "inside" into combinatorial discs, glued together along arcs in their boundary. So there's an inductive step where you have to prove that two discs glued together along arcs in their boundary is a disc. The problem with this is that you've got to use this inductive step many times, so you see the actual map you get may be quite complicated.

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  • $\begingroup$ "I believe the proof is fairly old". Well, yes, close to 150 years. See also the thread I linked to in a comment to the question. $\endgroup$ – t.b. Sep 4 '11 at 22:00
  • $\begingroup$ @Theo, I believe you mean this link: en.wikipedia.org/wiki/Schwarz%E2%80%93Christoffel_mapping and that's not a circle packing argument. The nice thing about the circle packing argument is it's pretty much already implemented in available software. $\endgroup$ – Ryan Budney Sep 4 '11 at 22:03
  • $\begingroup$ Yes, I meant the Schwarz-Christoffel formula -- I often forget to fix the dashes in Wiki-links. Yes, my mistake, I misinterpreted the meaning of "constructive". Never mind. Thanks for the clarification (and the link to Stephenson's book, I'll have to check it out, it looks beautiful). $\endgroup$ – t.b. Sep 4 '11 at 22:05
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    $\begingroup$ [7] 91a:53072 Gage, Michael E. Curve shortening on surfaces. Ann. Sci. École Norm. Sup. (4) 23 (1990), no. 2, 229--256. (Reviewer: Dennis M. DeTurck) 53C22 (35K55 58G11) $\endgroup$ – Ryan Budney Sep 5 '11 at 16:01
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    $\begingroup$ Another relevant reference: MR0979601 (90a:53050) Grayson, Matthew A.(1-UCSD) Shortening embedded curves. Ann. of Math. (2) 129 (1989), no. 1, 71–111. $\endgroup$ – Ryan Budney Sep 6 '11 at 18:53

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